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Suppose that $f:[0,1] \to \mathbb{R}$ is defined by $f(x) = 1$ when $x = \frac{1}{n}$ for some positive integer $n$ and $f(x) = 0$ otherwise. How can I prove that $f$ is Riemann integrable on $[0,1]$?

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Cover each $x = \frac1n$ by interval of size $\frac{\epsilon}{2^n}$ i.e. consider intervals around $\frac1n$ as $\left(\frac1n - \frac{\epsilon}{2^{n+1}}, \frac1n + \frac{\epsilon}{2^{n+1}} \right)$ –  user17762 Dec 9 '11 at 2:51
    
@Sivaram Perhaps I am missing something, but I am unsure how to proceed from your hint. Doesn't Riemann integration consider only finite partitions of $[0,1]$? –  Srivatsan Dec 9 '11 at 2:55
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@Srivatsan: This should go with the previous hint. Choose $N$ such that $\frac1N < \frac{\epsilon}{2}$. For all $n < N$, cover each $x = \frac1n$ by interval of size $\frac{\epsilon}{2^{n+1}}$ i.e. consider intervals around $\frac1n$ as $\left(\frac1n - \frac{\epsilon}{2^{n+2}}, \frac1n + \frac{\epsilon}{2^{n+2}} \right)$ and for the rest use an interval of the form $[0,\epsilon/2)$ –  user17762 Dec 9 '11 at 2:59
    
You need to use these intervals to show that the set of discontinuities has measure zero. math.ncku.edu.tw/~rchen/Advanced%20Calculus/… –  Henry Shearman Dec 9 '11 at 3:00
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2 Answers

Here's a more elementary argument. We show that for every tagged partition $P$ of $[0,1]$ of norm at most $\varepsilon \gt 0$, the lower and upper Riemann sums $L(f; P)$ and $U(f;P)$ of $f$ satisfy $$ 0 \leqslant L(f, P) \leqslant U(f;P) \leqslant 2\sqrt{\varepsilon} . \tag{1} $$ The Riemann integrability of $f$ follows as a consequence. The lower bound is obvious; we show only the upper bound of $O(\sqrt{\varepsilon})$ here.

Fix a parameter $t \in (0,1)$ (to be determined later). Without loss of generality With a modest loss in generality, assume that $t$ is the endpoints of two intervals in $P$. [I leave it to the OP to remove this assumption.] The strategy is to bound the contributions of the two types of intervals individually:

  • Those that fall to the left of $t$ (i.e., those that are a subset of $[0, t]$). The contribution from these intervals to the Riemann sum is at most $t$, the total length of the left part (since the function is always bounded above by $1$ in magnitude).

  • Those that fall to the right of $t$ (i.e., those that are a subset of $[t, 1]$). Of all such intervals, at most $1/t$ of them contain a point of the form $1/n$. Therefore, all but $1/t$ of them contribute zero to the Riemann sum. Since each of these intervals is of length at most $\varepsilon$, the total contribution of these to the Riemann sum is at most $\varepsilon \cdot \frac{1}{t} = \frac{\varepsilon}{t}$.

Adding the two contributions above, we get that the total Riemann sum is at most $t + \frac{\varepsilon}{t}$. Picking $t = \sqrt{\varepsilon}$ to optimise this upper bound, we get $(1)$.

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The discontinuities of the function are in the form $x=\frac{1}{n}$. Using the given hint by Sivaram Ambikasaran in the comments to construct covering intervals, you can show that the set

$$D=\left\{x\in[0,1]:x=\frac{1}{n}\right\}$$

has measure zero which satisfies the Lebesgue criterion for integrability.

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Applying the Lebesgue criterion for this problem seems like an overkill. :-) –  Srivatsan Dec 9 '11 at 3:01
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This is a nice and probably the efficient way to go provided the OP knows Lebesgue integral in the first place. –  user17762 Dec 9 '11 at 3:07
    
Probably is overkill, but the interval construction works very nicely in showing that the sets have measure zero. –  Henry Shearman Dec 9 '11 at 3:08
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The extra work to get a finite partition is so small, why not go ahead and do it? –  GEdgar Dec 9 '11 at 20:07
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